Advanced Higher Maths 2011-2012 : Discussion and Help

That's exactly what i did, i reasoned it would take forever to use conjugates...

15a 4
15b 7

16a 6
16b 4

...

Evan, can you add the usual how hard was this year compared to previous poll?

I have corrected the exampaper scan at
http://www.thestudentroom.co.uk/showpost.php?p=37685736&postcount=1495

(note UKdragon)

Ok do you want me to include it with the solutions in the same file or as the same post but two different files?

I'm looking at another hour or so perhaps before posting the solutions.

I think we all know what the answer to that will be.

What are you trying to say?

Wondered if someone could help me qith question 5, I want to know roughly how many marks I would get.

I made a mistake in finding the direction of QR and got (2,-2,1) instead of (2,-2,-2).

I am sure I done the normal vector then the equation of the plane right as well (ended up with 6x + 6y - 9z = 3), anyone know how many marks I would lose?

Well personally I found 2011 a lot easier. And talking to a bunch of people who have also done the 2011 paper, they found this year a lot harder.

I don't think you can get marks for consequential errors(except where its a completely separate part of the question where you have to use your answer to a previous part)


So only 1/5 I'm afraid

Well personally I found 2011 a lot easier. And talking to a bunch of people who have also done the 2011 paper, they found this year a lot harder.

i think it was really different in many ways to the past papers... definitely harder than last year.

Sorry, I know you're busy.

For the last question, everyone is saying to multiply by the complex conjugate (which I tried at first, but got nowhere ). Eventually, though, I started again and just changed the angles in the numerator to (pi/36) and changed the power to 22-I checked wolfram alpha and it says it's the same thing. I then cancelled out the denominator which left (cos(pi/36)+isin(pi/36))^18
Then, using de moivre's (cos on its own for quickness) cos(18pi/36) =cos pi/2=0?

Would this be alright?

Actually I just looked at the question and read through your solution. It looks completely fine and rather elegant! I think though I'll do it by a different method (and no, conjugates are not necessary).

For 16b) could you not just do (cos(pi/18)+isin(pi/18)^11 = cos(11pi/18)+isin(11pi/18) and (cos(pi/36)+isin(pi/36)^4 = cos(4pi/36)+isin(4pi/36) then when dividing just subtract

cos(11pi/18 - 2pi/18) + isin(11pi/18 - 2pi/18) = cis(9pi/18) = cos(pi/2)+isin(pi/2)

cos(pi/2) = 0

that for me is easiest and quickest way.

Actually I just looked at the question and read through your solution. It looks completely fine and rather elegant! I think though I'll do it by a different method (and no, conjugates are not necessary).

Have you seen some of the questions from the last 2 exams?


This year's questions seemed more conventional, bar the lack of a good old differential equation. Even the rate of change question was doable if I had not spit the dummy out.